Direction Vector Homework: Find Rates of Change in Celsius

[athrun200]

Homework Statement

2. The Celsius temperature in a region of space is given by T(x, y, z) = 2x^2 − xyz. A
particle is moving in this region with its position at time t given by x = 2t^2, y = 3t,
z = −t^2, where time and distance are measured in seconds and meters, respectively.
(a) Find the rate of change of the temperature experienced by the particle in C◦/m
when the particle is at the point P(8, 6,−4).
(b) Find the rate of change of the temperature experienced by the particle in C◦/sec
at P.

Homework Equations

The Attempt at a Solution

To find part a, we need to find the direction vector. However, how to find direction vector? Is velocity vector the direction vector?

If so, than part b is very easy.
Since $\frac{dT}{dt}$=$\frac{dT}{ds}\frac{ds}{dt}$
Where [/itex]=$\frac{ds}{dt}$ is the magnitute of velocity.

I wonder whether I am correct.

 

[HallsofIvy]

The "direction" vector is the unit tangent vector to the path. The "rate of change of T in degrees Celcius per meter" is $\nabla T\cdot \vec{v}\text{ }$ where $\vec{v}\text{ }$ is the unit tangent vector.

For part (b) at least, you do NOT need to find the "direction" or s, just use the chain rule:
$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial u}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$.

That is almost the same as (a) except that you are not dividing by the length of the tangent vector to get the unit tangent vector.

 

[athrun200]

The "direction" vector is the unit tangent vector to the path. The "rate of change of T in degrees Celcius per meter" is $\nabla T\cdot \vec{v}\text{ }$ where $\vec{v}\text{ }$ is the unit tangent vector.

For part (b) at least, you do NOT need to find the "direction" or s, just use the chain rule:
$$\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial u}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}$$.

That is almost the same as (a) except that you are not dividing by the length of the tangent vector to get the unit tangent vector.

That means we have 2 ways to do part b?
1. Is the chain rule.
2. Just like part (a) but without dividing the length of the tangent vector?

 

[HallsofIvy]

Yes, the derivative of < x, y, z> with respect to t gives a "tangent vector", dividing by its length gives the unit tangent vector. The chain rule essentiall is the dot product of grad T with that non-unit tangent vector.

Probably the simplest way to do this is to do (b) first, by the chain rule, then divide that answer by the length of the vector
$$(dx/dt)\vec{i}+ (dy/dt)\vec{j}+ (dz/dt)\vec{j}$$
to get the answer to (a). That is the magnitude of the velocity vector and so is the speed with which the particle is moving. "rate of change of T as a function of time" divided by "rate of change of position as a function of time" is "rate of change of T as a function of distance. In a very loose notation:
$$\frac{\frac{dT}{dt}}{\frac{dx}{dt}}= \frac{dT}{dt}\frac{dt}{dx}= \frac{dT}{dx}$$